A 67 kg soccer player jumps verticallyupwards and heads the 0.45 kg ball as it i

Question

A 67 kg soccer player jumps verticallyupwards and heads the 0.45 kg ball as it is descending verticallywith a speed of 25 m/s. If the player wasmoving upward with a speed of 4.0 m/s just before impact, what willbe the speed of the ball immediately after the collision if theball rebounds vertically upwards and the collision is elastic?

m/s

If the ball is in contact with the player's head for 20 ms, what is the average acceleration of the ball?(Note that the force of gravity may be ignored during the briefcollision time.)
m/s2

Explanation / Answer

Using conservation of momentum of the system (player +ball)      67kg x 4m/s - 0.45kg x 25m/s = 0 +0.45kg Vball   Remember that the ball is moving down before collision that iswhy its velocity is negative . Vball = 570.55   Vball = 571 m/s b) The impulse = Ft = change in momentum =p                           Ft = p2 - p1 = m (V2-V1)                           F = m/t (V2 -V1) = 0.45/20 x10-3 (570.55 -25) = 12.27 x103 N   Now,               F= ma                     12.27x103 N= 0.45 a                        a= 27.277 x10-3                      a = 0.027m/s2                                     Rate meplease

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